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Commit 1979699

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feat: add solutions to lc problems: No.2933,2934 (#1962)
* No.2933.High-Access Employees * No.2934.Minimum Operations to Maximize Last Elements in Arrays
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14 files changed

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‎solution/2900-2999/2933.High-Access Employees/README.md‎

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Original file line numberDiff line numberDiff line change
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<!-- 这里可写通用的实现逻辑 -->
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**方法一:哈希表 + 排序**
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我们用一个哈希表 $d$ 来存储每个员工的所有访问时间,其中键为员工的姓名,值为一个整数数组,表示该员工的所有访问时间,该时间为从当天 00:00 开始的分钟数。
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对于每个员工,我们将其所有访问时间按照从小到大的顺序进行排序。然后我们遍历该员工的所有访问时间,如果存在连续的三个访问时间 $t_1, t_2, t_3,ドル满足 $t_3 - t_1 < 60,ドル则该员工为高访问员工,我们将其姓名加入答案数组中。
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最后,返回答案数组即可。
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时间复杂度 $O(n \times \log n),ドル空间复杂度 $O(n)$。其中 $n$ 为访问记录的数量。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def findHighAccessEmployees(self, access_times: List[List[str]]) -> List[str]:
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d = defaultdict(list)
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for name, t in access_times:
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d[name].append(int(t[:2]) * 60 + int(t[2:]))
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ans = []
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for name, ts in d.items():
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ts.sort()
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if any(ts[i] - ts[i - 2] < 60 for i in range(2, len(ts))):
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ans.append(name)
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return ans
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public List<String> findHighAccessEmployees(List<List<String>> access_times) {
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Map<String, List<Integer>> d = new HashMap<>();
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for (var e : access_times) {
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String name = e.get(0), s = e.get(1);
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int t = Integer.valueOf(s.substring(0, 2)) * 60 + Integer.valueOf(s.substring(2));
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d.computeIfAbsent(name, k -> new ArrayList<>()).add(t);
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}
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List<String> ans = new ArrayList<>();
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for (var e : d.entrySet()) {
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String name = e.getKey();
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var ts = e.getValue();
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Collections.sort(ts);
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for (int i = 2; i < ts.size(); ++i) {
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if (ts.get(i) - ts.get(i - 2) < 60) {
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ans.add(name);
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break;
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}
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<string> findHighAccessEmployees(vector<vector<string>>& access_times) {
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unordered_map<string, vector<int>> d;
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for (auto& e : access_times) {
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auto name = e[0];
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auto s = e[1];
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int t = stoi(s.substr(0, 2)) * 60 + stoi(s.substr(2, 2));
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d[name].emplace_back(t);
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}
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vector<string> ans;
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for (auto& [name, ts] : d) {
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sort(ts.begin(), ts.end());
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for (int i = 2; i < ts.size(); ++i) {
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if (ts[i] - ts[i - 2] < 60) {
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ans.emplace_back(name);
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break;
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func findHighAccessEmployees(access_times [][]string) (ans []string) {
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d := map[string][]int{}
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for _, e := range access_times {
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name, s := e[0], e[1]
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h, _ := strconv.Atoi(s[:2])
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m, _ := strconv.Atoi(s[2:])
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t := h*60 + m
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d[name] = append(d[name], t)
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}
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for name, ts := range d {
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sort.Ints(ts)
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for i := 2; i < len(ts); i++ {
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if ts[i]-ts[i-2] < 60 {
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ans = append(ans, name)
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break
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}
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}
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}
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return
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}
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```
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### **TypeScript**
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```ts
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function findHighAccessEmployees(access_times: string[][]): string[] {
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const d: Map<string, number[]> = new Map();
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for (const [name, s] of access_times) {
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const h = parseInt(s.slice(0, 2), 10);
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const m = parseInt(s.slice(2), 10);
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const t = h * 60 + m;
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if (!d.has(name)) {
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d.set(name, []);
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}
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d.get(name)!.push(t);
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}
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const ans: string[] = [];
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for (const [name, ts] of d) {
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ts.sort((a, b) => a - b);
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for (let i = 2; i < ts.length; ++i) {
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if (ts[i] - ts[i - 2] < 60) {
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ans.push(name);
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break;
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}
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}
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}
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return ans;
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}
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```
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### **...**

‎solution/2900-2999/2933.High-Access Employees/README_EN.md‎

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@@ -59,30 +59,141 @@ So the answer is [&quot;ab&quot;,&quot;cd&quot;].</pre>
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## Solutions
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**Solution 1: Hash Table + Sorting**
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We use a hash table $d$ to store all access times of each employee, where the key is the employee's name, and the value is an integer array, representing all access times of the employee, which are the number of minutes from the start of the day at 00:00.
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For each employee, we sort all their access times in ascending order. Then we traverse all access times of the employee. If there are three consecutive access times $t_1, t_2, t_3$ that satisfy $t_3 - t_1 < 60,ドル then the employee is a high-frequency visitor, and we add their name to the answer array.
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Finally, we return the answer array.
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The time complexity is $O(n \times \log n),ドル and the space complexity is $O(n)$. Here, $n$ is the number of access records.
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<!-- tabs:start -->
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### **Python3**
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```python
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class Solution:
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def findHighAccessEmployees(self, access_times: List[List[str]]) -> List[str]:
79+
d = defaultdict(list)
80+
for name, t in access_times:
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d[name].append(int(t[:2]) * 60 + int(t[2:]))
82+
ans = []
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for name, ts in d.items():
84+
ts.sort()
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if any(ts[i] - ts[i - 2] < 60 for i in range(2, len(ts))):
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ans.append(name)
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return ans
6888
```
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### **Java**
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```java
73-
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class Solution {
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public List<String> findHighAccessEmployees(List<List<String>> access_times) {
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Map<String, List<Integer>> d = new HashMap<>();
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for (var e : access_times) {
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String name = e.get(0), s = e.get(1);
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int t = Integer.valueOf(s.substring(0, 2)) * 60 + Integer.valueOf(s.substring(2));
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d.computeIfAbsent(name, k -> new ArrayList<>()).add(t);
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}
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List<String> ans = new ArrayList<>();
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for (var e : d.entrySet()) {
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String name = e.getKey();
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var ts = e.getValue();
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Collections.sort(ts);
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for (int i = 2; i < ts.size(); ++i) {
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if (ts.get(i) - ts.get(i - 2) < 60) {
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ans.add(name);
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break;
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}
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}
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}
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return ans;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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vector<string> findHighAccessEmployees(vector<vector<string>>& access_times) {
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unordered_map<string, vector<int>> d;
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for (auto& e : access_times) {
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auto name = e[0];
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auto s = e[1];
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int t = stoi(s.substr(0, 2)) * 60 + stoi(s.substr(2, 2));
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d[name].emplace_back(t);
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}
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vector<string> ans;
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for (auto& [name, ts] : d) {
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sort(ts.begin(), ts.end());
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for (int i = 2; i < ts.size(); ++i) {
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if (ts[i] - ts[i - 2] < 60) {
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ans.emplace_back(name);
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break;
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}
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}
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}
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return ans;
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}
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};
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```
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### **Go**
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```go
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func findHighAccessEmployees(access_times [][]string) (ans []string) {
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d := map[string][]int{}
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for _, e := range access_times {
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name, s := e[0], e[1]
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h, _ := strconv.Atoi(s[:2])
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m, _ := strconv.Atoi(s[2:])
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t := h*60 + m
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d[name] = append(d[name], t)
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}
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for name, ts := range d {
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sort.Ints(ts)
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for i := 2; i < len(ts); i++ {
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if ts[i]-ts[i-2] < 60 {
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ans = append(ans, name)
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break
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}
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}
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}
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return
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}
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```
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### **TypeScript**
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```ts
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function findHighAccessEmployees(access_times: string[][]): string[] {
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const d: Map<string, number[]> = new Map();
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for (const [name, s] of access_times) {
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const h = parseInt(s.slice(0, 2), 10);
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const m = parseInt(s.slice(2), 10);
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const t = h * 60 + m;
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if (!d.has(name)) {
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d.set(name, []);
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}
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d.get(name)!.push(t);
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}
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const ans: string[] = [];
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for (const [name, ts] of d) {
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ts.sort((a, b) => a - b);
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for (let i = 2; i < ts.length; ++i) {
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if (ts[i] - ts[i - 2] < 60) {
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ans.push(name);
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break;
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}
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}
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}
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return ans;
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}
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```
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### **...**
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Original file line numberDiff line numberDiff line change
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class Solution {
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public:
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vector<string> findHighAccessEmployees(vector<vector<string>>& access_times) {
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unordered_map<string, vector<int>> d;
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for (auto& e : access_times) {
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auto name = e[0];
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auto s = e[1];
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int t = stoi(s.substr(0, 2)) * 60 + stoi(s.substr(2, 2));
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d[name].emplace_back(t);
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}
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vector<string> ans;
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for (auto& [name, ts] : d) {
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sort(ts.begin(), ts.end());
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for (int i = 2; i < ts.size(); ++i) {
15+
if (ts[i] - ts[i - 2] < 60) {
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ans.emplace_back(name);
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break;
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}
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}
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}
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return ans;
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}
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};
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Original file line numberDiff line numberDiff line change
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func findHighAccessEmployees(access_times [][]string) (ans []string) {
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d := map[string][]int{}
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for _, e := range access_times {
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name, s := e[0], e[1]
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h, _ := strconv.Atoi(s[:2])
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m, _ := strconv.Atoi(s[2:])
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t := h*60 + m
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d[name] = append(d[name], t)
9+
}
10+
for name, ts := range d {
11+
sort.Ints(ts)
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for i := 2; i < len(ts); i++ {
13+
if ts[i]-ts[i-2] < 60 {
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ans = append(ans, name)
15+
break
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}
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}
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}
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return
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}
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Original file line numberDiff line numberDiff line change
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class Solution {
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public List<String> findHighAccessEmployees(List<List<String>> access_times) {
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Map<String, List<Integer>> d = new HashMap<>();
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for (var e : access_times) {
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String name = e.get(0), s = e.get(1);
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int t = Integer.valueOf(s.substring(0, 2)) * 60 + Integer.valueOf(s.substring(2));
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d.computeIfAbsent(name, k -> new ArrayList<>()).add(t);
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}
9+
List<String> ans = new ArrayList<>();
10+
for (var e : d.entrySet()) {
11+
String name = e.getKey();
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var ts = e.getValue();
13+
Collections.sort(ts);
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for (int i = 2; i < ts.size(); ++i) {
15+
if (ts.get(i) - ts.get(i - 2) < 60) {
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ans.add(name);
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break;
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}
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}
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}
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return ans;
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}
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}
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class Solution:
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def findHighAccessEmployees(self, access_times: List[List[str]]) -> List[str]:
3+
d = defaultdict(list)
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for name, t in access_times:
5+
d[name].append(int(t[:2]) * 60 + int(t[2:]))
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ans = []
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for name, ts in d.items():
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ts.sort()
9+
if any(ts[i] - ts[i - 2] < 60 for i in range(2, len(ts))):
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ans.append(name)
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return ans

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